# Equivalent resistance in BJT small-signal : Rpi in parallel with Beta*Ib

I'm currently reading Vorperian's book on Fast Analytical Circuits techniques. On page 129 he states :

I'm not sure I understand correctly the statement I highlighted in yellow. Is it some kind of use of the Norton<->Thevenin conversion?

If anybody could make me understand what he means mathematically and/or schematically.

I know the resistance reflection rule in a BJT, but I'm not sure I understand it in terms of "parallel combination between $$\i_b \beta\$$ and $$\r_{\pi}\$$".

It doesn't seem the resulting equivalent circuit has $$\r_e\$$ in series with $$\R_E\$$ given eq. 4.59.

Any help will be appreciated. Thanks!

• – G36
Commented May 11, 2021 at 18:35

TThe dynamic resistance $$\r_e\$$ is nothing else than the resistance of the parallel combination of $$\r_{\pi}\$$ and the current source (nomenclature as in the figure):

Test voltage: $$\v_T=i_T r_e\$$ with $$\i_T=i_b + \beta i_b\$$ and with $$\i_b=\frac{v_T}{r_{\pi}}\$$.

From this you can solve for $$\r_e=\frac{r_{\pi}}{\beta+1}\$$

By the way (physical interpretation): the resistance $$\r_e\$$ is nothing else than the inverse transconductance $$\g_e\$$ for the transistor: $$\r_e=\frac{1}{g_e}=\frac{dV_{be}}{dI_e}\$$.

From the diagram you can derive that $$\r_e\$$ is in parallel with the sum of $$\R_E\$$ and $$\R_s||R_L\$$

• Hope you don't mind if I edited the equations with Mathjax; it is now easier to read your nice answer. Tschuess! Commented May 11, 2021 at 16:44
• Oh no - I do not mind. Just the opposite is true. Thank you so much.
– LvW
Commented May 11, 2021 at 17:47
• Thank you very much for the answer! Sorry for the delay :) Commented Jun 7, 2021 at 11:05